JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 1)
Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :
$$\frac{1}{2}$$
$$\frac{3}{2}$$
$$-$$$$\frac{3}{2}$$
$$\frac{5}{2}$$
Explanation
$${{xdy - ydx} \over {\sqrt {{x^2} + {y^2}} }} = dx$$
$$ \Rightarrow {{dy} \over {dx}} = {{\sqrt {{x^2} + {y^2}} } \over x} + {y \over x}$$
$$ \Rightarrow {{dy} \over {dx}} = \sqrt {1 + {{{y^2}} \over {{x^2}}}} + {y \over x}$$
Let $${y \over x} = v$$
$$ \Rightarrow v + x{{dv} \over {dx}} = \sqrt {1 + {v^2}} + v$$
$$ \Rightarrow {{dv} \over {\sqrt {1 + {v^2}} }} = {{dx} \over x}$$
OR $$\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C$$
at $$x = 1,\,y = 0$$
$$ \Rightarrow C = 0$$
$${y \over x} + \sqrt {1 + {{{y^2}} \over {{x^2}}}} = x$$
At $$x = 2$$,
$${y \over 2} + \sqrt {1 + {{{y^2}} \over 4}} = 2$$
$$ \Rightarrow 1 + {{{y^2}} \over 4} = 4 + {{{y^2}} \over 4} - 2y$$
OR $$y = {3 \over 2}$$
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